There are 2 answers to this question:
+ inspect the graph
+ use the quadratic formula
When I graph these numbers:
I get a graph showing that at x = 0.0625, y is less than 50; at x = 0.125, y is greater than 50. So, I expect the value of x giving y = 50 to be between 0.0625 and 0.125; and, somewhat closer to 0.125 than to 0.0625.
Fitting these data to a 2nd-order polynomial, I get
y = -203.05x2 + 109.13x + 40.239
The quadratic formula is devised to give 2 x values (possibly imaginery) that will make y = 0. To find x values that make y = 50, I need to change the quadratic I am solving for.
50 = -203.05x^2 + 109.13x + 40.239
0 = -203.05x^2 + 109.13x - 9.761
So,
x = ( -109 +/- sqrt(109^2 - 4*203*9.761)) / (2 * 203)
For the first solution:
x = ( -109.13 + SQRT(109.13^2 - 4*203.05*9.761)) / (2 * -203.05)
x = 0.1133491
For the second solution:
x = ( -109.13 - SQRT(109.13^2 - 4*203.05*9.761)) / (2 * -203.05)
x = 0.4241
Check my numbers, I might be starting with the wrong data, or making a mistake somewhere.
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